Use the Euler-Lagrange equations (motivated in a previous post of mine). These are
$$ \partial_x L - \frac{\mathrm{d}}{\mathrm{d}t} \left( \partial_{v_x} L \right) = 0 $$
We therefore need
$$ \partial_x L = -q \partial_x \Phi + q \left( \partial_x A_x v_x + \partial_x A_y v_y + \partial_x A_z v_z \right) $$
$$ \partial_{v_x} L = m v_x + q A_x $$
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left( \partial_{v_x} L \right) = m \dot v_x + q \dot A_x $$
Putting this all together into the first equation and adding $\dot v_x m$ gives the equation for which you are looking.
Note especially that each component of $A$ is a function of all three coordinates: $A_i = A_i(x,y,z)$; the same goes for $\Phi = \Phi(x,y,z)$ (where usually $A_0$ is identified with $\Phi$ in relativistic electrodynamics).
Goldstein states
Both $\mathbf{E}(t,x,y,z)$ and $\mathbf{B}(t,x,y,z)$ are continuous functions of time and position derivable from a scalar potential $\phi(t,x,y,z)$ and a vector potential $\mathbf{A}(t,x,y,z)$
i.e., these terms are not dependent on $\mathbf{v}$.
Thus,
$$
\frac{\partial\phi}{\partial\dot{x}}=\frac{\partial\mathbf{A}}{\partial\dot{x}}=0
$$
and, trivially, the time derivatives are also zero. The only term that depends on $\dot{x}$ is the kinetic energy term, hence the appearance of $m\ddot{x}$ and the disappearance of your term.
With regards to interchanging derivatives, with an ordinary derivative you can write it as
$$
\frac{df}{dt} = \sum_{n=1}^p\frac{\partial f}{\partial q_n}\dot{q}_n+\frac{\partial f}{\partial t}
$$
for any $f(q_1,\ldots,q_p,\dot{q}_1,\ldots,\dot{q}_p,t)$. If you multiply the above by $\partial/\partial \dot{q}_i$ and carefully apply partial derivatives, you will actually find
$$
\frac{\partial}{\partial\dot{q}_i}\left(\frac{df}{dt}\right) = \frac{d}{dt}\left(\frac{\partial f}{\partial \dot{q}_i}\right)+\frac{\partial f}{\partial q_i}
$$
(in your case, $\dot{q}_i=\dot{x}$).
For partial derivatives, it should be straight-forward that
$$
\frac{\partial^2f}{\partial x\partial y}=\frac{\partial^2f}{\partial y\partial x}
$$
Best Answer
Velocity-dependent potential is not strictly a potential. Lagrange equations say that
$$\frac{d}{dt}\frac{\partial L}{\partial \bf{v}} = \frac{\partial L}{\partial \bf{r}}$$
You have $L = L_0 - U$ where $L_0$ corresponds to free motion (e.g. $L_0 = mv^2/2$ or $L_0 = -mc^2\sqrt{1-(v/c)^2}$).
If $U$ does not contain $\bf{v}$ you have ${\partial L}/{\partial \bf{v}} = \bf{p}$ and so $\dot{\bf{p}} = -\nabla U$.
In this case, however $U$ contains $\bf{v}$ so on the you have
$$\frac{d}{dt}\left({\bf{p}} + q\bf{A}\right) = -\nabla U$$